学生成绩表(stuscore):
姓名:name
课程:subject
分数:score
学号:stuid
张三
数学
89
1
张三
语文
80
1
张三
英语
70
1
李四
数学
90
2
李四
语文
70
2
李四
英语
80
2
创建表 SET ANSI_NULLS ONGOSET QUOTED_IDENTIFIER ONGOSET ANSI_PADDING ONGOCREATE TABLE [dbo].[stuscore]( [name] [varchar](50) COLLATE Chinese_PRC_CI_AS NULL, [subject] [varchar](50) COLLATE Chinese_PRC_CI_AS NULL, [score] [int] NULL, [stuid] [int] NULL) ON [PRIMARY] GOSET ANSI_PADDING OFF
问题:
1. 计算每个人的总成绩并排名(要求显示字段:姓名,总成绩)
2. 计算每个人的总成绩并排名(要求显示字段: 学号,姓名,总成绩)
3. 计算每个人单科的最高成绩(要求显示字段: 学号,姓名,课程,最高成绩)
4. 计算每个人的平均成绩(要求显示字段: 学号,姓名,平均成绩)
5. 列出各门课程成绩最好的学生(要求显示字段: 学号,姓名,科目,成绩)
6. 列出各门课程成绩最好的两位学生(要求显示字段: 学号,姓名,科目,成绩)
7. 统计如下:
学号
姓名
语文
数学
英语
总分
平均分
8.列出各门课程的平均成绩(要求显示字段:课程,平均成绩)
9.列出数学成绩的排名(要求显示字段:学号,姓名,成绩,排名)
10.列出数学成绩在2-3名的学生(要求显示字段:学号,姓名,科目,成绩)
11.求出李四的数学成绩的排名
12.统计如下:
课程
不及格(0-59)个
良(60-80)个
优(81-100)个
13.统计如下:数学:张三(50分),李四(90分),王五(90分),赵六(76分)
答案:
12. 课程 不及格(-59) 良(-80) 优(-100)
select subject, ( select count ( * ) from stuscore where score < 60 and subject = t1.subject) as 不及格,( select count ( * ) from stuscore where score between 60 and 80 and subject = t1.subject) as 良,( select count ( * ) from stuscore where score > 80 and subject = t1.subject) as 优 from stuscore t1 group by subject 13. 数学: 张三 (50 分 ), 李四 (90 分 ), 王五 (90 分 ), 赵六 (76 分 ) declare @s varchar ( 1000 ) set @s = '' select @s = @s + ' , ' + name + ' ( ' + convert ( varchar ( 10 ),score) + ' 分) ' from stuscore where subject = ' 数学 ' set @s = stuff ( @s , 1 , 1 , '' ) print ' 数学: ' + @s
1. 计算每个人的总成绩并排名
select name, sum (score) as allscore from stuscore group by name order by allscore2. 计算每个人的总成绩并排名
select distinct t1.name,t1.stuid,t2.allscore from stuscore t1,( select stuid, sum (score) as allscore from stuscore group by stuid)t2 where t1.stuid = t2.stuid order by t2.allscore desc3. 计算每个人单科的最高成绩
select t1.stuid,t1.name,t1.subject,t1.score from stuscore t1,( select stuid, max (score) as maxscore from stuscore group by stuid) t2 where t1.stuid = t2.stuid and t1.score = t2.maxscore4.计算每个人的平均成绩
select distinct t1.stuid,t1.name,t2.avgscore from stuscore t1,( select stuid, avg (score) as avgscore from stuscore group by stuid) t2 where t1.stuid = t2.stuid5.列出各门课程成绩最好的学生
select t1.stuid,t1.name,t1.subject,t2.maxscore from stuscore t1,( select subject, max (score) as maxscore from stuscore group by subject) t2 where t1.subject = t2.subject and t1.score = t2.maxscore6.列出各门课程成绩最好的两位学生
select distinct t1. * from stuscore t1 where t1.stuid in ( select top 2 stuscore.stuid from stuscore where subject = t1.subject order by score desc ) order by t1.subject
7.学号 姓名 语文 数学 英语 总分 平均分
select stuid as 学号,name as 姓名, sum ( case when subject = ' 语文 ' then score else 0 end ) as 语文, sum ( case when subject = ' 数学 ' then score else 0 end ) as 数学, sum ( case when subject = ' 英语 ' then score else 0 end ) as 英语, sum (score) as 总分,( sum (score) / count ( * )) as 平均分 from stuscore group by stuid,name order by 总分desc
8.列出各门课程的平均成绩
select subject, avg (score) as avgscore from stuscore group by subject9.列出数学成绩的排名
declare @tmp table (pm int ,name varchar ( 50 ),score int ,stuid int ) insert into @tmp select null ,name,score,stuid from stuscore where subject = ' 数学 ' order by score desc declare @id int set @id = 0 ; update @tmp set @id = @id + 1 ,pm = @id select * from @tmpselect DENSE_RANK () OVER ( order by score desc ) as row,name,subject,score,stuid from stuscore where subject = ' 数学 ' order by score desc
declare @tmp table (pm int identity ( 1 , 1 ),name varchar ( 50 ),score int ,stuid int ) insert into @tmp select name,score,stuid from stuscore where subject = ' 数学 ' order by score desc select * from @tmp
10. 列出数学成绩在2-3名的学生
select t3. * from ( select top 2 t2. * from ( select top 3 name,subject,score,stuid from stuscore where subject = ' 数学 ' order by score desc ) t2 order by t2.score) t3 order by t3.score desc11. 求出李四的数学成绩的排名
declare @tmp table (pm int ,name varchar ( 50 ),score int ,stuid int ) insert into @tmp select null ,name,score,stuid from stuscore where subject = ' 数学 ' order by score desc declare @id int set @id = 0 ; update @tmp set @id = @id + 1 ,pm = @id select * from @tmp where name = ' 李四 '